week 2 of mth 209. due for this week… homework 2 (on mymathlab – via the materials link) ...

Week 2 of MTH 209

Due for this week…

Homework 2 (on MyMathLab – via the Materials Link) Monday night at 6pm.

Read Chapter 6.6-6.7, 7.6-7.7, 10.5, 11.3-11.4 Do the MyMathLab Self-Check for week 2. Learning team planning for week 5. Discuss your final week topic for your team

presentations…

Slide 2Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Introduction to Factoring

Common Factors

Factoring by Grouping

6.1


Common Factors

When factoring a polynomial, we first look for factors that are common to each term.

By applying a distributive property we can often write a polynomial as a product.

For example: 8x2 = 2x 4x and 6x = 2x 3

And by the distributive property,8x2 + 6x = 2x(4x + 3)


EXAMPLE

Solution

Finding common factors

Factor. a. b. c. d.

a.

26 7x x

d.

3 215 5x x 3 28 2 4w w w 2 3 2 26x y x y

b.

c.

26 7x x 26 6

7 7

(6 7)

x x

x

x

x

x x

3 215 5x x 3

2

2

2

2

15 3

5

(3 1)

5

5

5

x x

x

x

x

x

x

3 28 2 4w w w 3 2

2

2

8 4

2

4 2

2 (4

2

2

)

2

2

w w

w w

w

w

ww

w w w

2 3 2 26x y x y2 3

2 2

2

2 2

2 2

26 6

(6 1)

x y

x

x y

x y y

x y y


When finding the greatest common factor for a polynomial, it is often helpful first to completely factor each term of the polynomial.


EXAMPLE

Solution

Finding the greatest common factor

Find the greatest common factor for the expression. Then factor the expression. 18x2 + 3x

18x2 = 2 3 3 x x

3x = 3 x

The greatest common factor is 3x.

18x2 + 3x = 3x(6x + 1)


Factoring by Grouping

Factoring by grouping is a technique that makes use of the associative and distributive properties.


EXAMPLE

Solution

Factoring out binomials

Factor. a. 3x(x + 1) + 4(x + 1) b. 3x2(2x – 1) – x(2x – 1)

a. Both terms in the expression contain the binomial x + 1. Use the distributive property to factor.

1( ) 4 )1(3 xx x ( )( 13 4 )x x

b. 23 (2 1) (2 1)x x x x 2( ) (2 )1 13 2x xx x

2( ) 2 1)3 (x xx ( )(2 11 )3x xx


EXAMPLE

Solution

Factoring by grouping when the middle term is (+)

Factor the polynomial.

3 24 3 12x x x

3 24 3 12x x x 3 2( 4 ) (3 12)x x x

3 2( 4 ) (3 12)x x x 2( ) 3( )4 4x x x

2( )( )3 4x x


EXAMPLE

Solution

Factoring by grouping when the middle term is ()

Factor the polynomial. 3 215 10 3 2x x x

3 215 10 3 2x x x 3 2(15 10 ) ( 3 2)x x x 2( )3 2 3 2)5 1( xx x 2( )(35 1 )2x x


EXAMPLE

Solution

Factoring out the GCF before grouping

Completely factor the polynomial. 3 230 20 6 4x x x

3 230 20 6 4x x x

3 22[(15 10 ) ( 3 2)]x x x

2 3 22[ ( ) ( )]3 25 1xx x

25 12( )( )3 2x x

3 22(15 10 3 2)x x x

Practice for section 6.1

Finding common factors questions 13-20 Finding the greatest common factor Q 21-38 Factoring out binomials Q 39-44 Factoring by grouping when the middle term is (+)Q 45-48 Factoring by grouping when the middle term is ()Q 53-59 Factoring out the GCF before grouping Q 65-70



Factoring Trinomials I (x2 + bx + c)

Review of the FOIL Method

Factoring Trinomials Having a Leading Coefficient of 1

6.2


The product (x + 3) (x + 4) can be found as follows:x2 + 4x + 3x + 12x2 + 7x + 12

The middle term is found by calculating the sum 4x and 3x, and the last term is found by calculating the product of 4 and 3.

To solve equations using factoring, we use the zero-product property. It states that, if the product of two numbers is 0, then at least one of the numbers must equal 0.


EXAMPLE

Solution

Factoring a trinomial having only positive coefficients

Factor each trinomial. a. b.

a.

2 11 18x x 2 10 24x x

b. 2 11 18x x

( 2)( 9)x x

Factors of 18 whose sum is 11.

Factors Sum

1, 18 19

2, 9 11

3, 6 9

2 10 24x x Factors of 24 whose sum is 10.

Factors Sum

1, 24 25

2, 12 14

3, 8 11

4, 6 10

( 4)( 6)x x


EXAMPLE

Solution

Factoring a trinomial having a negative middle coefficient


a.

2 9 14x x 2 19 48x x

b.

( 2)( 7)x x

Factors of 14 whose sum is −9.

Factors Sum

−1, −14 −15

−2, −7 −9

Factors of 48 whose sum is −19.

Factors Sum

−1, −48 −49

−2, −24 −26

−3, −16 −19

−4, −12 −16

−6, −8 −14

( 3)( 16)x x

2 9 14x x 2 19 48x x


EXAMPLE

Solution

Factoring a trinomial having a negative constant term


a.

2 2 15x x 2 6 16x x

b.

( 3)( 5)x x

Factors of −15 whose sum is 2.

Factors Sum

1, −15 −14

− 1, 15 +14

3, −5 −2

−3, 5 2

Factors of −16 whose sum is −6.

Factors Sum

−1, 16 15

1, −16 −15

−2, 8 6

2, −8 −6

−4, 4 0

( 2)( 8)x x

2 2 15x x 2 6 16x x


EXAMPLE

Solution

Discovering that a trinomial is prime

Factor the trinomial. 2 6 8x x

Factors of −8 whose sum is 6.

Factors Sum

1, −8 −7

−1, 8 7

2, −4 −2

−2, 4 2

2 6 8x x

The table reveals that no such factor pair exists. Therefore, the trinomial is prime.


EXAMPLE

Solution

Factoring out the GCF before factoring further

Factor each trinomial completely. a. b.

a.

24 28 48x x 27 21 70x x

b. 24 28 48x x

4( 3)( 4)x x

Factor out common factor of 4.

Factors (12) Sum (7)

1, 12 13

2, 6 8

3, 4 7

27 21 70x x Factor out common factor of 7.

Factors (10) Sum (3)

1, 10 9

2, 5 3

2, 5 3

7( 2)( 5)x x

24( 7 12)x x 27( 3 10)x x


EXAMPLE

Solution

Finding the dimensions of a rectangle

Find one possibility for the dimensions of a rectangle that has an area of x2 + 12x + 35.

The area of a rectangle equals length times width. If we can factor x2 + 12x + 35, then the factors can represents its length and width.

x

x 5

7 35

5x

7x

x2

Because x2 + 12x + 35 = (x + 5)(x + 7), one possibility for the rectangle’s dimensions is width x + 5 and length x + 7.

Area = x2 + 12x + 35


Factoring a trinomial having only positive coefficients Q 15-26

Factoring a trinomial having a negative middle coefficient Q 29-38

Factoring a trinomial having a negative constant term Q 39-60

Factoring out the GCF before factoring further Q 61-80

Finding the dimensions of a rectangle Q 83



Factoring Trinomials II (ax2 + bx + c)

Factoring Trinomials by Grouping

Factoring with FOIL in Reverse

6.3


Factoring Trinomials by Grouping


EXAMPLE

Solution

Factoring ax2 + bx + c by grouping


a. b.

212 5 3y y 22 13 15x x

22 13 15x x Multiply (2)(15) = 30Factors of 30 whose sum is 1310 and 3

2 32 1510x xx 2(2 10 ) (3 15)x x x

( )52 ( )3 5x xx ( )(2 )3 5x x

212 5 3y y Multiply (12)(3) = 36Factors of 36 whose sum is 5 9 and 4

2 912 34y yy 2(12 ) ( 39 4 )yy y

( )4 3 4 3)3 1( yy y ( )(4 31 )3y y


EXAMPLE

Solution

Discovering that a trinomial is prime

Factor the trinomial.

a. We need to find integers m and n such that mn = (3)(4) = 12 and m + n = 9. Because the middle term is positive, we consider only positive factors of 12.

23 9 4x x

There are no factors whose sum is 9, the coefficient of the middle term. The trinomial is prime.

Factors 1, 12 2, 6 3, 4

Sum 13 8 7


Factoring with FOIL in Reverse


EXAMPLE

Solution

Factoring the form ax2 + bx + c


22 7 6x x

22 7 6x x 22 7 6 (2 ___)( ___)x x x x

The factors of the last term are either 1 and 6 or 2 and 3.Try a set of factors. Try 1 and 6.

2(2 1)( 6) 2 13 6

12

13

x

x

x x x x

x

Middle term is 13x not 7x.


EXAMPLE

Solution

Factoring the form ax2 + bx + c—continued


22 7 6x x

2(2 2)( 3) 2 8 6

8

2

6

x

x

x x x x

x

Try 2 and 3 the factors of the last term.

Middle term is 8x not 7x.

Try another set of factors 3 and 2.

2(2 3)( 2) 2 7 6

3

4

7

x

x

x x x x

x

Middle term is correct.

The factors of the last term are either 1 and 6 or 2 and 3.Try a set of factors.


Factoring ax2 + bx + c by grouping Q11-34 Discovering that a trinomial is prime Q 13-23 Factoring the form ax2 + bx + c Q 29-50



Special Types of Factoring

Difference of Two Squares

Perfect Square Trinomials

Sum and Difference of Two Cubes

6.4


For any real numbers a and b, a2 – b2 = (a – b)(a + b).

DIFFERENCE OF TWO SQUARES


EXAMPLE

Solution

Factoring the difference of two squares

a. 9x2 – 16 b. 5x2 + 8y2 c. 25x4 – y6

a. 9x2 – 16 = (3x)2 – (4)2 = (3x – 4) (3x + 4)

b. Because 5x2 + 8y2 is the sum of two squares, it cannot be factored.

c. If we let a2 = 25x4 and b2 = y6, then a = 5x2 and b = y3. Thus,

25x4 – y6 = (5x2)2 – (y3)2

= (5x2 – y3) (5x2 + y3).

Factor each difference of two squares.


For any real numbers a and b, a2 + 2ab + b2 = (a + b)2 and a2 − 2ab + b2 = (a − b)2.

PERFECT SQUARE TRINOMIALS


EXAMPLE

Solution

Factoring perfect square trinomials

a. x2 + 8x + 16 b.

a.

If possible, factor each trinomial as a perfect square.

Let a2 = x2 and b2 = 42. For a perfect square trinomial, the middle term must be 2ab.

2ab = 2(x)(4) = 8x,

which equals the given middle term. Thus a2 + 2ab + b2 = (a + b)2 implies

x2 + 8x + 16 = (x + 4)2.

4x2 − 12x + 9

x2 + 8x + 16


EXAMPLE continued

b.

Let a2 = (2x)2 and b2 = 32. For a perfect square trinomial, the middle term must be 2ab.

2ab = 2(2x)(3) = 12x,

which equals the given middle term. Thus a2 − 2ab + b2 = (a − b)2 implies

4x2 − 12x + 9 = (2x − 3)2.

4x2 − 12x + 9


For any real numbers a and b, a3 + b3 = (a + b)(a2 – ab + b2) and a3 − b3 = (a − b)(a2 + ab + b2)

SUM AND DIFFERENCE OF TWO CUBES


EXAMPLE

Solution

Factoring the sum and difference of two cubes

a. n3 + 27 b.

a.

Factor each polynomial.

Because n3 = (n)3 and 27 = 33, we let a = n, b = 3, and factor. Substituting a3 + b3 = (a + b)(a2 – ab + b2) gives n3 + 33 = (n + 3)(n2 – n ∙ 3 + 32) = (n + 3)(n2 – 3n + 9).

8x3 − 125y3

n3 + 27


EXAMPLE continued

b.

Here, 8x3 = (2x)3 and 125y3 = (5y)3, so 8x3 − 125y3 = (2x)3 – (5y)3.

Substituting a = 2x and b = 5y in a3 − b3 = (a − b)(a2 + ab + b2)

gives (2x)3 – (5y)3 = (2x − 5y)(4x2 + 10xy +

25y2).

8x3 − 125y3


EXAMPLE

Solution

Factoring out the GCF before factoring further

8s3 – 32st2Factor the polynomial completely.

Factor out the common factor of 8s.

8s3 – 32st2 = 8s(s2 – 4t2)

= 8s(s – 2t)(s + 2t)


Factoring the difference of two squares Q 15-30 Factoring perfect square trinomials Q 31-52 Factoring the sum and difference of two cubes

Q53-66 Factoring out the GCF before factoring further

Q 67-84



Introduction to Rational Expressions

Basic Concepts

Simplifying Rational Expressions

Applications

7.1


Rational expressions can be written as quotients (fractions) of two polynomials.Examples include:

2

3

2

, , 5

4

6

3

4 1

4 8

x x

x

x

xx

Basic Concepts


EXAMPLE

Solution

Evaluating rational expressions

If possible, evaluate each expression for the given value of the variable.a. b. c.

a.

1; 3

3x

x

2

; 43 4

ww

w

4; 5

4

ww

w

b. c. 1; 3

3x

x

2

;3 4

4w

ww

4;

45

ww

w

1

3

1

3 6

2( )

3

4

( 44)

16

12 4

16

28

4 ( )

5)

5

( 4

91

9


EXAMPLE

Solution

Determining when a rational expression is undefined

Find all values of the variable for which each expression is undefined.a. b. c.

a.

2

1

x

2

4

w

w 2

6

4w

b. c. 2

1

x

2

4

w

w 2

6

4w

Undefined when x2 = 0 or when x = 0.

Undefined when w – 4 = 0 or when w = 4.

Undefined when w2 – 4 = 0 or when w = 2.


EXAMPLE

Solution

Simplifying fractions

Simplify each fraction by applying the basic principle of fractions.a. b. c.

a. The GCF of 9 and 15 is 3.

9

15

20

2845

135

9

15 5

3

3

3

3

5

b. 20

28 7

4

4

5

The GCF of 20 and 28 is 4.

5

7

c. The GCF of 45 and 135 is 45. 45

135

145

5 34

1

3


EXAMPLE

Solution

Simplifying rational expressions

Simplify each expression.a. b. c.

a.

2

16

4

y

y

3 12

4 16

x

x

2

2

25

2 7 15

x

x x

b. c. 2

16

4

y

y

3 12

4 16

x

x

44

4

y

y y

4

y

3( )

4(

4

)4

x

x

3

4

2

2

25

2 7 15

x

x x

( )( 5)

(2

5

3)( )5

x

x

x

x

5

2 3

x

x


EXAMPLE

Solution

Distributing a negative sign

Simplify each expression.a. b.

a.

7

2 14

y

y

8

8

x

x

b. 7

2 14

y

y

8

8

x

x

1( )

2(

7

7)

y

y

1

2

(8 )

8

x

x

8

8

x

x

8

18

x

x


Evaluating rational expressions Q 11-24 Determining when a rational expression is undefined

Q 29-42 Simplifying fractions Q 43-50 Simplifying rational expressions Q 55-66,71-84 Distributing a negative sign Q 67-70



Multiplication and Division of Rational Expressions

Review of Multiplication and Division of Fractions

Multiplication of Rational Expressions

Division of Rational Expressions

7.2


EXAMPLE

Solution

Multiplying fractions

Multiply and simplify your answers to lowest terms.

a. b. c.

4 5

9 7

a.

b.

415

5 2 5

7 8

4 5 20

9 7 63

4 15 4 6015 12

5 1 5 5

c.2 5 5 2 5 1 5

7 8 7 8 7 4 28


EXAMPLE

Solution

Dividing fractions

Divide and simplify your answers to lowest terms.

a. b. c.

1 3

6 5

a.

b.

618

7 4 11

5 15

c.

1 3

6 5

618

7

4 11

5 15

1 5

6 3

5

18

1 6

7 18

6 1

7 18

1

21

60

55

4 15

5 11

12

11

12 5

11 5


EXAMPLE

Solution

Multiplying rational expressions

Multiply and simplify to lowest terms. Leave your answers in factored form. a. b.

2

6 5

10 12

x

x

3 4

2 1 3 9

x x

x x

a.b.

2

6 5

10 12

x

x

210 12

6 5x

x

21

0

0

3

2

x

x

1

4x

3 4

2 1 3 9

x x

x x

( 3)( 4)

(2 1) 3 )9(x x

x x

( )( 4)

3(2 1)( 3)

3 x

x

x

x

4

3(2 1)

x

x


EXAMPLE

Solution

Multiplying rational expressions

Multiply and simplify to lowest terms. Leave your answer in factored form.

2

2

16 3

9 4

x x

x x

2

2

( 3)

(

( )

4)

6

( 9)

1

x

x

x

x

2

2

16 3

9 4

x x

x x

( 3)( 4)( 4)

( 3)( 3 4))(x

xx x

x x

( 4)( )( )

( 3)

3

( 3 )

4

4)(

x

x

x x

x x

( 4)

( 3)

x

x


EXAMPLE

Solution

Dividing rational expressions

Divide and simplify to lowest terms.a. b. 3 2 1

6

x

x x

2

2

16 4

2 8 2

x x

x x x

a. b.3 2 1

6

x

x x

3 6

2 1

x

x x

18

(2 1)

x

x x

18

2 1x

2

2

16

2 8

4

2

x

x x

x

x

2

2

16

42 8

2x

x x

x

x

( 4)( 4) 2

( 2)( 4) 4

x x x

x x x

( )( )( )

( )( )(

4

2 4

24

)4

x

xx

x

x

x

1


Multiplying fractions Q 9-14 Dividing fractions Q 17-22 Multiplying rational expressions Q 33-49 Dividing rational expressions Q 53-70



Addition and Subtraction With Like Denominators

Review of Addition and Subtraction of Fractions

Rational Expressions Having Like Denominators

7.3


EXAMPLE

Solution

Adding fractions having like denominators

Add and simplify to lowest terms.

a. b.

a.

4 1

7 7

1 5

9 9

4 1

7 7 4 1

7

5

7

b. 1 5

9 9 1 5

9

6

9

2

3


EXAMPLE

Solution

Subtracting fractions having like denominators

Subtract and simplify to lowest terms.

a. b.

a.

13 7

18 18

15 11

30 30

13 7

18 18

13 7

18

6

18

1

3

b. 15 11

30 30

15 11

30

2

15

4

30


To add two rational expressions having like denominators, add their numerators. Keep the same denominator.

C is nonzero

SUMS OF RATIONAL EXPRESSIONS

A B A B

C C C


EXAMPLE

Solution

Adding rational expressions having like denominators

Add and simplify.

a. b.

a.

4 1 2

3 3

x x

x x

2 2

5

7 10 7 10

x

x x x x

b.

3

4

3

1 2

x x

x x

4 1 2

3

x x

x

3

5 1x

x

2 2

5

7 10 7 10

x

x x x x

2

5

7 10

x

x x

5

5 2

x

x x

1

2x


EXAMPLE

Solution

Adding rational expressions having two variables

Add and simplify to lowest terms.

a. b.

a.

7 4

ab ab 2 2 2 2

w y

w y w y

b.

7 4

ab ab

7 4

ab

11

ab

2 2 2 2

w y

w y w y

2 2

w y

w y

( )( )

w y

w y w y

1

w y


To subtract two rational expressions having like denominators, subtract their numerators. Keep the same denominator.

C is nonzero

DIFFERENCES OF RATIONAL EXPRESSIONS

A B A B

C C C


EXAMPLE

Solution

Subtracting rational expressions having like denominators


a. b.

a.

2 2

6 6x

x x

2 2

2 3 4

1 1

x x

x x

b.

2 2

6 6x

x x

2

6 6x

x

2

6 6x

x

2

x

x

1

x

2 2

2 3 4

1 1

x x

x x

2

2 3 4

1

x x

x

1

1 1

x

x x

2

1

1

x

x

1

1x


EXAMPLE

Solution

Subtracting rational expressions having like denominators


7 2

2 2

a a

a a

7 2

2 2

a a

a a

7 ( 2)

2

a a

a

6 2

2

a

a


Adding fractions having like denominators Q 15-20 Subtracting fractions having like denominators

Q 27-34 Adding rational expressions having like

denominators Q 25, 39, 41 Adding rational expressions having two variables

Q 57-62 Subtracting rational expressions having like

denominators Q 37, 63



Addition and Subtraction with Unlike Denominators

Finding Least Common Multiples

Review of Fractions Having Unlike Denominators

Rational Expressions Having Unlike Denominators

7.4


The least common multiple (LCM) of two or more polynomials can be found as follows.

Step 1: Factor each polynomial completely. Step 2: List each factor the greatest number of

times that it occurs in either factorization.Step 3: Find the product of this list of factors. The

result is the LCM.

FINDING THE LEAST COMMON MULTIPLE


EXAMPLE

Solution

Finding least common multiples

Find the least common multiple of each pair of expressions.

Factor each polynomial completely.

a. 6x, 9x4

Step 1:

b. x2 + 7x + 12, x2 + 8x + 16

a.

6x = 3 ∙ 2 ∙ x 9x4 = 3 ∙ 3 ∙ x ∙ x ∙ x ∙ x

Step 2: List each factor the greatest number of times.

3 ∙ 3 ∙ 2 ∙ x ∙ x ∙ x ∙ x

Step 3: The LCM is 18x4.


EXAMPLE continued

Factor each polynomial completely.Step 1:

b.

Step 2: List each factor the greatest number of times.

Step 3: The LCM is (x + 3)(x + 4)2.

x2 + 7x + 12, x2 + 8x + 16

x2 + 7x + 12 = (x + 3)(x + 4)

x2 + 8x + 16 = (x+ 4)(x + 4)

(x + 3), (x + 4), and (x + 4)

Find the least common multiple.


EXAMPLE

Solution

Adding and subtracting fractions having unlike denominators

Simplify each expression.

a. b.

a. The LCD is the LCM, 42.

4 1

7 6

4 1

7 6

4 6 1 7

7 6 6 7 24 7

42 42

31

42

5 11

12 30

b. The LCD is 60.5 11

12 30

5 5 11 2

12 5 30 2

25 22

60 60

3

60

1

20


EXAMPLE

Solution

Adding rational expressions having unlike denominators

Find each sum and leave your answer in factored form.

a. b.

a.

2

2 5

x x

4 3

1 1x x

b.

2

2 5

x x

2

2 5x

x x x

2 2

2 5x

x x

2

2 5x

x

4 3

1 1x x

4 3 1

1 1 1x x

4 3

1 1x x

1

1x


EXAMPLE

Solution

Subtracting rational expressions having unlike denominators

Simply each expression. Write your answer in lowest terms and leave it in factored form.

The LCD is x(x + 7).

3 5

7

x

x x

3 5

7

x

x x

3 7 5

7 7

x x x

x x x x

3 7 5

7 7

x x x

x x x x

3 7 5

7

x x x

x x

2 4 21 5

7

x x x

x x

2 21

7

x x

x x


EXAMPLE Subtracting rational expressions with unlike denominators

The LCD is (x + 3)(x + 3)(x – 3).

2 2

6 5

6 9 9x x x

6 5

3 3 3 3x x x x

3 36 5

3 3 3 3 3 3

x x

x x x x x x

6 3 5 3

3 3 3 3 3 3

x x

x x x x x x

6 18 5 15

3 3 3

x x

x x x

33

3 3 3

x

x x x

Simplify each expression. Write your answer in lowest terms and leave it in factored form.


EXAMPLE

Solution

Modeling electrical resistance

Add and then find the reciprocal of the result.

The LCD is RS.

1 1,

R S

1 1 1 1

R S R

R

S

S

S R

S R

RS RS

S R

RS

The reciprocal is

.RS

S R


Finding least common multiples Q 15-38 Adding and subtracting fractions having unlike

denominators Q 51-58 Adding rational expressions having unlike

denominators Q 59-77 Subtracting rational expressions having unlike

denominators Q 69-87 Modeling electrical resistance Q 103



Radical Expressions and Rational Exponents

Radical Notation

Rational Exponents

Properties of Rational Exponents

10.1


Every positive number a has two square roots, one positive and one negative. Recall that the positive square root is called the principal square root.

The symbol is called the radical sign.

The expression under the radical sign is called the radicand, and an expression containing a radical sign is called a radical expression.

Examples of radical expressions:

57, 6 2, and

3 4

xx

x


EXAMPLE Finding principal square roots

Evaluate each square root.

a.

b.

c.

36

0.64

4

5

0.8

16

25

6


EXAMPLE Approximating a square root

Approximate to the nearest thousandth. 38

6.164


EXAMPLE Finding nth roots

Find each root, if possible. a. b. c.

Solutiona.

4 256 5 243 4 1296

b.

c.

4 256

5 243

4 1296

4 because 4 4 4 4 256.

53 because ( 3) 243.

An even root of a negative number is not a real number.


EXAMPLE Simplifying expressions

Write each expression in terms of an absolute value. a. b. c.

Solutiona.

2( 5) 2( 3)x 2 6 9w w

b.

c.

2( 5)

2( 3)x

2 6 9w w

5 5

3x

2( 3)w 3w


EXAMPLE Interpreting rational exponents

Write each expression in radical notation. Then evaluate the expression and round to the nearest hundredth when appropriate. a. b. c.

Solution

a.

1/249 1/526 1/26x

b.

c.

1/249 1/526

1/2(6 )x

49

6x

7


EXAMPLE Interpreting rational exponents

Write each expression in radical notation. Evaluate the expression by hand when possible. a. b.

Solutiona.

3/481 4/514

b. /4381 4/514

43/(81)

34 81

Take the fourth root of 81 and then cube it.

33

27

Take the fifth root of 14 and then fourth it.

54/14

45 14

Cannot be evaluated by hand.


EXAMPLE Interpreting negative rational exponents

Write each expression in radical notation and then evaluate. a. b.

Solutiona.

1/481 2/364

b. 1/481 2/3641/4

1

81

4

1

81

1

3

3/2

1

64

23

1

64

2

1

4

1

16


EXAMPLE Applying properties of exponents

Write each expression using rational exponents and simplify. Write the answer with a positive exponent. Assume that all variables are positive numbers.

a. b. Solution

a.

4x x 34 256x

b. 4x x 34 256x1/2 1/4x x

1/2 1/4x

3/4x

1/43(256 )x

1/4 3 1/4256 ( )x

3/44x


EXAMPLE Applying properties of exponents-continued

c. d.

Solution

a.

5

4

32x

x

1/33

27

x

b.

5

4

32x

x

1/5

1/4

(32 )x

x

1/5 1/5

1/4

32 x

x

1/5 1/42x

1/3

3

27

x

1/3

3 1/3

27

( )x

3

x1/202x

1/20

2

x

1/33

27

x


EXAMPLE Applying properties of exponents

Write each expression with positive rational exponents and simplify, if possible.

a. b.

Solution

a.

4 2x

1/4

1/5

y

x

b. 1/41/2( 2)x

1/8( 2)x

1/5

1/4

x

y

4 2x 1/4

1/5

y

x


Finding principal square roots Q 11-18 Approximating a square root Q 35-36 Finding nth roots Q 29-34 Simplifying expressions Q 103-110 Interpreting rational exponents Q 41-56,63-68 Interpreting negative rational exponents Q 57-62 Applying properties of exponents Q 91-102



Simplifying Radical Expressions

Product Rule for Radical Expressions

Quotient Rule for Radical Expressions

10.2


Consider the following example:

4 25 2 5 10

4 25 100 10

Note: the product rule only works when the radicals have the same index.


EXAMPLE Multiplying radical expressions

Multiply each radical expression.

a.

b.

c.

36 4

3 38 27

4 41 1 1 1 1

4 16 4 256 4

3 38 27 216 6

4 441 1 1

4 16 4

36 4 144 12

Solution


EXAMPLE Multiplying radical expressions containing variables

Multiply each radical expression. Assume all variables are positive.

a.

b.

c.

2 4x x

3 23 5 10a a

44 4

3 7 2121

x y xy

y x xy

3 32 3 35 10 50 50a a a a

443 7x y

y x

2 4 6 3x x x x

Solution


EXAMPLE Simplifying radical expressions

Simplify each expression. a. b. c.

Solutiona.

500 3 40 72

b.

c.

500

3 40

72

100 5 10 5

3 3 38 5 2 5

36 2 6 2



Simplify each expression. Assume that all variables are positive. a. b. c.

Solution

a.

449x 575y 3 23 3 9a a w

b. c.

449x

575y

4 249 7x x

4 325y y 3 23 9a a w

4 325y y

25 3yy

3 23 3 9a a w

3 327a w

3 33 27a w 33a w


EXAMPLE Multiplying radicals with different indexes

Simplify each expression.

a. b.

Solution

37 7 3 5a a

a. b. 37 7 1/2 1/37 7 1/3 1/5a 1/2 1/37

5/67

3 5a a

8/15a

1/3 1/5a a


Consider the following examples of dividing radical expressions: 4 2 2 2

9 3 3 3

4 4 2

9 39


EXAMPLE Simplifying quotients

Simplify each radical expression. Assume that all variables are positive. a. b.

Solution

a.

37

275

32

x

b. 37

27

3

3

7

27

3 7

3

5

32

x 5

5 32

x

5

2

x


EXAMPLE Simplifying quotients

Simplify each radical expression. Assume that all variables are positive. a. b.

Solution

a.

90

10

4x y

y

b. 90

10

90

10

9

4x y

y

4x y

y

4x3 2x



Simplify the radical expression. Assume that all variables are positive.

Solution

4

55

32x

y

4

55

32x

y

5 4

55

32x

y

5 45

55

32 x

y

5 42 x

y


Multiplying radical expressions Q 11-22 Multiplying radical expressions containing

variables Q 23-62 (not 33, 39,41) Simplifying radical expressions Q 75-80 Multiplying radicals with different indexes

Q 101-110 Simplifying quotients Q 33,39,41 Simplifying radical expressions Q 95-98



Operations on Radical Expressions

Addition and Subtraction

Multiplication

Rationalizing the Denominator

10.3


Like radicals have the same index and the same radicand.

Like Unlike 3 2 5 2 3 2 5 3


EXAMPLE Adding like radicals

If possible, add the expressions and simplify.

a.

b.

c.

d.

4 7 8 7

3 37 5 2 5

6 16

39 5

8 13

12 7

Solution

The terms cannot be added because they are not like radicals.

The expression contains unlike radicals and cannot be added.


EXAMPLE Finding like radicals

Write each pair of terms as like radicals, if possible.

a. b. 80, 125 3 34 16, 7 54

125 25 5 5 5

80 16 5 4 5 33 34 16 4 8 2

33 37 54 7 27 2

3 34 2 2 8 2

3 37 3 2 21 2

Solutiona. b.


EXAMPLE Adding radical expressions

Add the expressions and simplify.

a. b. 20 5 5 5 2 50 72

4 5 5 5 20 5 5 5 2 50 72

=5 2 5 2 6 2

5 2 25 2 36 2

16 2

Solutiona.

2 55 5

57

b.


EXAMPLE Subtracting like radicals

Simplify the expressions.

a.

b.

8 7 2 7

3 3 37 5 2 5 5 3(7 2 1) 5

6 7

Solution

36 5


EXAMPLE Subtracting radical expressions

Subtract and simplify. Assume that all variables are positive. a. b.

Solution

5 549x x3

377

64 4

yy

a. b.

4 449x x x x

02 27x x x x

26x x

3 37 7

4 4

y y

5 549x x3

377

64 4

yy


EXAMPLE Subtracting radical expressions

Subtract and simplify. Assume that all variables are positive. a. b.

Solution

a.

7 2 3 2

5 3

3 7 4 3343 24 27a b ab

b. 7 2 3 2

5 3

7 2 3 3 2 5

5 3 3 5

21 2 15 2

15 15

6 2 2 2

15 5

3 7 4 3343 27a b ab

3 6 3 3 3 3343 27a b ab ab

2 3 37 3a b ab ab

2 3(7 3)a b ab


EXAMPLE Multiplying radical expressions

Multiply and simplify.

Solution

3 5x x

3 5x x 3 5 3 5x x x x

215 3 5x x x

15 2 x x


EXAMPLE Rationalizing the denominator

Rationalize each denominator. Assume that all variables are positive.a. b. c.

Solution

1

3

7

8 35

ab

b

a.1

33

3

3

3

b.7

8 33

3 7 3

8 3

7 3

24

c.5

ab

b2

ab

b b

2

ab b

b b b

2

ab b

b b

2

a b

b


Examples of conjugates.


EXAMPLE Using a conjugate to rationalize the denominator

Rationalize the denominator of

Solution

1.

1 3

1

1 31 3

1 33 1

1

2 2

1 3

1 ( 3)

1 3

1 3

1 3

2

1 3

2 2

1 3

2 2


EXAMPLE Rationalizing the denominator

Rationalize the denominator.

Solution

4 6

3 6

3 6

3 6

4 6

3 6

2

2

12 4 6 3 6 ( 6)

9 ( 6)

18 7 6

3

18 7 6

3 3

7 66

3

4 6

3 6


EXAMPLE Rationalizing the denominator having a cube root

Rationalize the denominator.

Solution

3 2

4

x

2/3

4

x

1/3

2/3 1/3

4 x

x x

1/3

2/3 1/3

4x

x

34 x

x

3 2

4

x

Practice for section 10.3 Adding like radicals Q 19-30 Finding like radicals Q 9-18 Adding radical expressions Q 29-49 Subtracting like radicals Q 33-40 Subtracting radical expressions Q 55-76 Multiplying radical expressions Q 77-88 Rationalizing the denominator Q 89-98 Using a conjugate to rationalize the denominator

Q 99-102 Rationalizing the denominator Q 103-108 Rationalizing the denominator having a cube root

Q 113-116



Complex Numbers

Basic Concepts

Addition, Subtraction, and Multiplication

Powers of i

Complex Conjugates and Division

10.6


PROPERTIES OF THE IMAGINARY UNIT i

21 and 1i i

THE EXPRESSION a

If > 0, then = . a a i a


EXAMPLE

Solution

Writing the square root of a negative number

Write each square root using the imaginary i.

36a. b. c.15 45

a. 36 36i 6i

b. 15 15i

c. 45 45i 9 5i 3 5i


Let a + bi and c + di be two complex numbers. Then

(a + bi) + (c+ di) = (a + c) + (b + d)i Sum

and(a + bi) − (c+ di) = (a − c) + (b − d)i. Difference

SUM OR DIFFERENCE OF COMPLEX NUMBERS


EXAMPLE

Solution

Adding and subtracting complex numbers

Write each sum or difference in standard form.

a. (−8 + 2i) + (5 + 6i) b. 9i – (3 – 2i)

a.

b.

(−8 + 2i) + (5 + 6i) = (−8 + 5) + (2 + 6)i = −3 + 8i

9i – (3 – 2i) = 9i – 3 + 2i = – 3 + (9 + 2)i = – 3 + 11i


EXAMPLE

Solution

Multiplying complex numbers

Write each product in standard form.

a. (6 − 3i)(2 + 2i) b. (6 + 7i)(6 – 7i)

a. (6 − 3i)(2 + 2i)

= (6)(2) + (6)(2i) – (2)(3i) – (3i)(2i)

= 12 + 12i – 6i – 6i2

= 12 + 12i – 6i – 6(−1)

= 18 + 6i


EXAMPLE continued

b. (6 + 7i)(6 − 7i)

= (6)(6) − (6)(7i) + (6)(7i) − (7i)(7i)

= 36 − 42i + 42i − 49i2

= 36 − 49i2

= 36 − 49(−1)

= 85


The value of in can be found by dividing n (a positive integer) by 4. If the remainder is r, then

in = ir.

Note that i0 = 1, i1 = i, i2 = −1, and i3 = −i.

POWERS OF i


EXAMPLE

Solution

Calculating powers of i

Evaluate each expression

a. i25 b. i7 c. i44

a. When 25 is divided by 4, the result is 6 with the remainder of 1. Thus i25 = i1 = i.

b. When 7 is divided by 4, the result is 1 with the remainder of 3. Thus i7 = i3 = −i.

c. When 44 is divided by 4, the result is 11 with the remainder of 0. Thus i44 = i0 = 1.


EXAMPLE

Solution

Dividing complex numbers

Write each quotient in standard form.

a. b.

a.

3 2

5

i

i

9

3i

3 2

5

i

i

3 2 5

5 5

i i

i i

3 5 3 2 5 2

5 5 5 5

i i i i

i i i i

2

2

15 3 10 2

25 5 5

i i i

i i i

15 7 2 1

25 1

i

17 7

26

i

17 7

26 26

i


EXAMPLE continued

b. 9

3i

9 3

3 3

i

i i

2

27

9

i

i

27

9 1

i

27

9

i

3i

Practice for section 10.6 Writing the square root of a negative number

Q 13-22 Adding and subtracting complex numbers

Q 23-30 Multiplying complex numbers Q 31-36 Calculating powers of i Q 49-56 Dividing complex numbers Q 71-74


MTH 209 End of week 2

You again have the answers to those problems not assigned

Practice is SOOO important in this course. Work as much as you can with MyMathLab, the

materials in the text, and on my Webpage. Do everything you can scrape time up for, first the

hardest topics then the easiest. You are building a skill like typing, skiing, playing a

game, solving puzzles. NEXT TIME: Nonlinear Equations – quadratic

equations, proportion and variation problems

week 2 of mth 209. due for this week… homework 2 (on mymathlab – via the materials link) ...

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